The `K_(sp)` of `Mg(OH)_(2)` is `1xx10^(-12). 0.01M Mg^(2+)` will precipitate tate at the limiting pH of

Options

A. `1.3`
B. `9`
C. `3.5`
D. `8`

Answer

B

Explanation

`Mg(OH)_(2)(s)hArr Mg^(2+)(aq.)+2OH^(-)(aq.)`
`:. K_(sp)= C_(Mg^(2+))C_(OH^(-))^(2)`
`10^(-12)=(10^(-2))C_(OH^(-))^(2)`
or `C_(OH^(-))^(2)=10^(-10)`
`C_(OH^(-))=10^(-5)M`
For any aqueous solution at `25^(@)C`,
`C_(H^(+))C_(OH^(-))=10^(-14)`
Hence, `C_(H^(+))=(10^(-14))/(C_(OH^(-)))=(10^(-14))/(10^(-5))=10^(-9)`
Since precipitation occurs when ionic product just exceeds `K_(sp)`, we have the limiting pH of `9(C_(H^(+))=10^(-pH)mol L^(-))`.